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Question Number 108825 by ajfour last updated on 19/Aug/20

Σ_(n=1) ^(10) (i^n +i^(n+1) )= ?

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right)=\:? \\ $$

Commented by ajfour last updated on 19/Aug/20

thanks everyone who confirmed.

$${thanks}\:{everyone}\:{who}\:{confirmed}. \\ $$

Answered by john santu last updated on 19/Aug/20

   ((⊸JS⊸)/♥)  Σ_(n=1) ^(10) i^n (1+i)=(1+i)Σ_(n=1) ^(10) i^n = (1+i)(i−1)=−1−1=−2  Σ_(k=1) ^(10) i^n  = i−1−i+1+i−1−i+1+i−1              = i−1(∗)

$$\:\:\:\frac{\multimap{JS}\multimap}{\heartsuit} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{i}^{{n}} \left(\mathrm{1}+{i}\right)=\left(\mathrm{1}+{i}\right)\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{i}^{{n}} =\:\left(\mathrm{1}+{i}\right)\left({i}−\mathrm{1}\right)=−\mathrm{1}−\mathrm{1}=−\mathrm{2} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{i}^{{n}} \:=\:{i}−\mathrm{1}−{i}+\mathrm{1}+{i}−\mathrm{1}−{i}+\mathrm{1}+{i}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:{i}−\mathrm{1}\left(\ast\right) \\ $$$$ \\ $$

Commented by ajfour last updated on 19/Aug/20

Σ_(k=1) ^(10) i^n =i−1−i+1+i−1−i+1+i−1             = i−1  so   Σ^(10) _(n=1) i^n (1+i)=(i+1)(i−1)                            = −1−1=−2 .  isn′t this right?

$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{i}^{{n}} ={i}−\mathrm{1}−{i}+\mathrm{1}+{i}−\mathrm{1}−{i}+\mathrm{1}+{i}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{i}−\mathrm{1} \\ $$$${so}\:\:\:\underset{{n}=\mathrm{1}} {\sum}^{\mathrm{10}} {i}^{{n}} \left(\mathrm{1}+{i}\right)=\left({i}+\mathrm{1}\right)\left({i}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\mathrm{1}−\mathrm{1}=−\mathrm{2}\:. \\ $$$${isn}'{t}\:{this}\:{right}? \\ $$

Commented by bemath last updated on 19/Aug/20

yes

$${yes} \\ $$

Commented by john santu last updated on 19/Aug/20

yes

$${yes} \\ $$

Answered by Dwaipayan Shikari last updated on 19/Aug/20

i((i^(10) −1)/(i−1))+i^2 ((i^(10) −1)/(i−1))=i(i+1)−(i+1)=−1−1=−2

$${i}\frac{{i}^{\mathrm{10}} −\mathrm{1}}{{i}−\mathrm{1}}+{i}^{\mathrm{2}} \frac{{i}^{\mathrm{10}} −\mathrm{1}}{{i}−\mathrm{1}}={i}\left({i}+\mathrm{1}\right)−\left({i}+\mathrm{1}\right)=−\mathrm{1}−\mathrm{1}=−\mathrm{2} \\ $$

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